Question

prove that the radius of a circle which circumscribes an equilateral triangle of side 12 cm is equal to 4√3 cm.​

Answer 1

Answer:

Let D be the mid point of a side AB of the equilateral triangle.

Now consider the triangle AOB in which angle

AOB = 2 angle ACB

as angle subtended by the chord AB at the centre  is twice the angle subtended by AB at C on the opposite part of circumference. So angle AOB = 120 degree.

Triangle OAD and OBD are congruent as

OD is common

ODA = ODB = 90°  

AB= BD ( as the the line joining the mid point of the chord and the centre is perpendicular to the chord. So Angle AOD = Angle BOD = 120/2 = 60.)

Therefore 2OD =  OB  = R the radius  Or

OD = R/2.

QB = R.

DB = 12/2 = 6cm

Applying Pythagoras theorem,

OD² + DB² = OD².

(R/2)² + 6² = R²

6² = R² - (R/2)²

6² = (3/4)R²

R² = (4 X 6²) / 3

R² = 48

R = 4√3

Hence Proved

Answer 2

hence proved that the radius of the circle is 4√3